Projectile Motion of a Cannonball: Calculating Velocity, Height, Time, and Range

In summary: If the ground is not level, then the landing direction will mirror the launch direction and the landing velocity will have a different direction, but the same magnitude as the launch velocity.I completely agree with you from beginning to the end. I will remind myself not to confused with standard-form equation.In summary, the conversation discussed the physics of projectile motion, specifically a cannonball fired at an angle of 25.0° above the horizontal with an initial velocity of 125 m/s. The horizontal and vertical components of the initial velocity were determined to be 113.3 m/s and 52.8 m/s, respectively. The maximum height of the cannonball was calculated to be 142.4 m, with a time of 5.4
  • #1
Glenboro
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Homework Statement


A cannonball is fired with a velocity of 125 m/s at 25.0° above the horizontal

a) Determine the horizontal and vertical components of the initial velocity.
b) Determine the maximum height the cannonball reaches in its path
c) Determine the time it takes to reach maximum height.
d) Determine the hang time of the cannonball.
e) Determine the range of the cannonball.
f) Draw a sketch of the motion of the projectile including the horizontal and vertical components of velocity when the projectile strikes the ground. Draw the appropriate relative lengths of the vectors for these components.

Homework Equations


a) V1x = v1 (cos), V1y = v1 (sin)
b) v2^2 = v1 + 2ad
c) D= (0.5)(V1 + V2)(t)
d) t = (v2- v1)/a
e) dx = (V1x)(t)
f) unknown

The Attempt at a Solution



a) V1x = (125 m/s)(cos 25) = 113.3 m/s right
V1y = (125 m/s)(sin 25) = 52.8 m/s up
b) (0m/s)^2 - (52.8m/s)^2/ (2 x -9.80 m/s2) = 142.4 m
c) 142.2 = (0.5)(52.8 m/s +0 m/s)(t)
142.2 = 26.4 x t
T = 5.4 s
d) T = [0 m/s)-(125 m/s)]/ -9.8m/s^2 = 12.76
12.76 x 2= 25.52 s (“Hang time” refers to the length of time that a projectile is in the air.)
e) dx = (113.3 m/s)(25.52s)
= 2.89 x 10^4m
f) If I know which equation I should use for this one, I will appreciate it.[/B]
 
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  • #2
You appear to be working by "remembering the right equation" then putting numbers into it. This is not a good way to do physics.
Instead of trying to remember the right equation, or get someone to tell you what it is, why not use your understanding of the physics involved to help you?

(d) curious - I always thought that "hang time" was the time when the ball appears stationary or in level flight near the top of it's trajectory. But I had a quick look and you are correct - I always used to call this "time of flight".

(f) no equation: just draw a picture using your understanding of how the ball moves.
 
  • #3
Simon Bridge said:
You appear to be working by "remembering the right equation" then putting numbers into it. This is not a good way to do physics.
Instead of trying to remember the right equation, or get someone to tell you what it is, why not use your understanding of the physics involved to help you?

(d) curious - I always thought that "hang time" was the time when the ball appears stationary or in level flight near the top of it's trajectory. But I had a quick look and you are correct - I always used to call this "time of flight".

(f) no equation: just draw a picture using your understanding of how the ball moves.
Yeah, that is a big problem for me. I'm more focused on getting numbers into equation rather than trying to put into my head. I will fix that habit asap.
Also, yes hang time is the time when projectile is in the air. Do you recognize any an error from my calculations?
 
  • #4
I don't check arithmetic. By the time you are studying projectile motion you should be confident you can get that right.
When you write equations, it is best to use the same variable to refer to the same thing and different variables to refer to different things. Resist the urge to blindly follow standard-form equations.
i.e. you know the max height h is reached when the upwards speed is zero, so, from the suvat equations, or from a vy-t graph: $$h=\frac{v^2\sin^2\theta}{2g}$$ ... this sort of writing makes your reasoning clear, and that is what I check.

You should know the general shape of a projectile's trajectory - just sketch that shape and label the important points ... just like when you sketch graphs in maths class.
They also want you to draw in the velocity components ... you should be able to see how they should go once you've sketched the basic shape.

Sketching graphs is one of the most important skills you can learn - you should also learn to use v-t graphs: fewer equations to memorize that way.
 
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  • #5
Simon Bridge said:
I don't check arithmetic. By the time you are studying projectile motion you should be confident you can get that right.
When you write equations, it is best to use the same variable to refer to the same thing and different variables to refer to different things. Resist the urge to blindly follow standard-form equations.
i.e. you know the max height h is reached when the upwards speed is zero, so, from the suvat equations, or from a vy-t graph: $$h=\frac{v^2\sin^2\theta}{2g}$$ ... this sort of writing makes your reasoning clear, and that is what I check.

You should know the general shape of a projectile's trajectory - just sketch that shape and label the important points ... just like when you sketch graphs in maths class.
They also want you to draw in the velocity components ... you should be able to see how they should go once you've sketched the basic shape.

Sketching graphs is one of the most important skills you can learn - you should also learn to use v-t graphs: fewer equations to memorize that way.
I completely agree with you from beginning to the end. I will remind myself not to confused with standard-form equation.
 
  • #6
heres a crib sheet i have:
Note: if the ground is level, then the landing direction and velocity will mirror the launch.
 

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